Problem 58 - Intermediate Algebra

Question

The graph of equation is an ellipse. Determine which distance is longer, the distance between the \(x\)-intercepts or the distance between the \(y\) -intercepts. How much longer? \(x^{2}+4 y^{2}=36\)

Step-by-Step Solution

Answer

The x-intercepts are 6 units longer than the y-intercepts.

Step 1: Rewrite the Equation in Standard Form

The given equation is \( x^2 + 4y^2 = 36 \). Divide every term by 36 to rewrite it in standard form of an ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We get: \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \). This corresponds to \( a^2 = 36 \) and \( b^2 = 9 \), so \( a = 6 \) and \( b = 3 \).

Step 2: Determine the x-Intercepts

To find the \( x \)-intercepts, set \( y = 0 \) in the standard form of the ellipse: \( \frac{x^2}{36} = 1 \). Solving for \( x \), we get \( x^2 = 36 \), so \( x = \pm 6 \). The \( x \)-intercepts are \( (6, 0) \) and \( (-6, 0) \).

Step 3: Calculate the Distance Between x-Intercepts

The distance between the points \( (6, 0) \) and \( (-6, 0) \) on the x-axis is \( 6 - (-6) = 12 \).

Step 4: Determine the y-Intercepts

To find the \( y \)-intercepts, set \( x = 0 \) in the standard form of the ellipse: \( \frac{y^2}{9} = 1 \). Solving for \( y \), we get \( y^2 = 9 \), so \( y = \pm 3 \). The \( y \)-intercepts are \( (0, 3) \) and \( (0, -3) \).

Step 5: Calculate the Distance Between y-Intercepts

The distance between the points \( (0, 3) \) and \( (0, -3) \) on the y-axis is \( 3 - (-3) = 6 \).

Step 6: Compare the Distances and Find the Difference

The distance between the \( x \)-intercepts is 12, and the distance between the \( y \)-intercepts is 6. The distance between the \( x \)-intercepts is longer by \( 12 - 6 = 6 \).

Key Concepts

Standard Form of an Ellipse

In geometry, the equation of an ellipse can take a specific structure known as the standard form, which simplifies the way we analyze the ellipse's properties. The standard form of an ellipse is expressed as: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]Here, the values of \( a^2 \) and \( b^2 \) are essential in determining the dimensions of the ellipse:

• \( a \) represents the semi-major axis length if \( a > b \), indicating the longer radius.
• \( b \) represents the semi-minor axis length if \( b < a \), indicating the shorter radius.

To convert the given equation \( x^2 + 4y^2 = 36 \) into its standard form:

• Divide every term by 36 to isolate the ellipse's equation: \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \).
• Identify that \( a^2 = 36 \) and \( b^2 = 9 \), thus \( a = 6 \) and \( b = 3 \).

This transformation clarifies the ellipse's orientation and size, helping us find intercepts and distances.

x-intercepts

Finding the x-intercepts of an ellipse is like locating where the ellipse crosses the x-axis. For this, we set \( y = 0 \) in the standard form equation, which means solving:\[ \frac{x^2}{36} = 1 \]This equation simplifies to:

• \( x^2 = 36 \)
• \( x = \pm 6 \)

Therefore, the ellipse crosses the x-axis at the points \((6, 0)\) and \((-6, 0)\). This means when you observe the graph of the ellipse, these x-intercepts represent the furthest left and right points on the x-axis. These points are crucial for geometric insights into the ellipse and help in further calculations, like finding distances.

y-intercepts

Just as we find the x-intercepts by setting \( y = 0 \), to find the y-intercepts, we set \( x = 0 \) in our ellipse's equation, discovering where it crosses the y-axis. This process involves:\[ \frac{y^2}{9} = 1 \]Upon solving this, it simplifies to:

• \( y^2 = 9 \)
• \( y = \pm 3 \)

Thus, the y-intercepts occur at the points \((0, 3)\) and \((0, -3)\). These intercepts denote the uppermost and lowermost parts of the ellipse along the y-axis, marking how the ellipse stretches vertically. Understanding these y-intercepts is key to calculating vertical distances and further analyzing the ellipse's features.

Distance Calculation

To fully understand the dimensions of an ellipse, we calculate the distances between the intercepts along each axis. This highlights the width and height of the ellipse:

• **Distance Between x-Intercepts:** - The points \((6, 0)\) and \((-6, 0)\) lie on the x-axis. - The distance formula in one-dimensional terms calculates as: \(|6 - (-6)| = 12\).
• **Distance Between y-Intercepts:** - The points \((0, 3)\) and \((0, -3)\) are on the y-axis. - Similarly, the distance is \(|3 - (-3)| = 6\).

These distances are quickly obtainable because of the symmetry of the ellipse. By comparing these distances, we determine that the x-intercept span (12) is longer by 6 than the y-intercept span (6). This difference in distances provides insight into the ellipse being elongated horizontally.